Design a data structure that supports all following operations in average O(1) time.
insert(val): Inserts an item val to the set if not already present.
remove(val): Removes an item val from the set if present.
getRandom: Returns a random element from current set of elements. Each element must have the same probability of being returned.
Example:
// Init an empty set.
RandomizedSet randomSet = new RandomizedSet();
// Inserts 1 to the set. Returns true as 1 was inserted successfully.
randomSet.insert(1);
// Returns false as 2 does not exist in the set.
randomSet.remove(2);
// Inserts 2 to the set, returns true. Set now contains [1,2].
randomSet.insert(2);
// getRandom should return either 1 or 2 randomly.
randomSet.getRandom();
// Removes 1 from the set, returns true. Set now contains [2].
randomSet.remove(1);
// 2 was already in the set, so return false.
randomSet.insert(2);
// Since 2 is the only number in the set, getRandom always return 2.
randomSet.getRandom();
解答:
1.rand()%nums.size() 是取随机数方法,从0到noms.size()-1;
2.设一个map和一个vector的原因是这样可以很方便地取随机数。map里存的是nums里val的位置。nums存的是所有val的取值。
class RandomizedSet {
public:
/** Initialize your data structure here. */
RandomizedSet() {
}
/** Inserts a value to the set. Returns true if the set did not already contain the specified element. */
bool insert(int val) {
if(m.find(val)!=m.end()) return false;
nums.push_back(val);
m[val]=nums.size()-1 ;
return true;
}
/** Removes a value from the set. Returns true if the set contained the specified element. */
bool remove(int val) {
if(m.find(val)==m.end()) return false;
int last=nums.back();
nums[m[val]]=last;
m[last]=m[val];
nums.pop_back();
m.erase(val);
return true;
}
/** Get a random element from the set. */
int getRandom() {
return nums[rand()%nums.size()];
}
private:
vector<int> nums;
map<int,int> m;
};
/**
* Your RandomizedSet object will be instantiated and called as such:
* RandomizedSet obj = new RandomizedSet();
* bool param_1 = obj.insert(val);
* bool param_2 = obj.remove(val);
* int param_3 = obj.getRandom();
*/